Consider a Vernier calipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale .In the Vernier calipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:

Vernier Calipers                                                            Screw Gauge 1MSD=18cm                                                         Pitch of the screw  P=214MSD = 5 VSD                                                               P = 2HSD⇒1VSD=45MSD                                          Least count =P100=150HSD⇒1VSD=45×18cm=110cm=1mm Least count = 1MSD – 1VSD⇒LC=18−110cm=140cm=14mm (B)  Pitch of the screw gauge is twice the least count of the Vernier calipers P=2HSD=14mm2 ⇒1HSD=14mmLeast count of screw Gauge  =15014mm=0.005mm(C) L.C of linear scale of screw gauge = 1HSDGiven that 1HSD =  214mm=12mmL.C of screw gauge  =150HSD=15012mm=0.01mm